If the Quadratic equation $p(x)=0$ with real coefficient has purely Imaginary roots.Then the
equation $p(p(x)) = 0$ has
$\bf{OPTIONS::}$
$(a)\;\; $ Only purely Inaginary Roots. $\;\;\;\;\;\;(b)$ all real roots.
$(c)$ Two real and Two purely imaginary roots.$\;\;\;\;\;\;(d)$ neither real nor purely imaginary roots.
$\bf{My\; Try}::$ Let $p(x) = x^2+1.$ Then $p(p(x)) = (x^2+1)^2+1 = x^4+2x^2+2$.
Now Roots of $\displaystyle p(p(x))=0\Rightarrow x^4+2x^2+2=0\Rightarrow x^2 = \frac{-2\pm\sqrt{4-8}}{2} = \frac{-2\pm 2i}{2} = -1\pm i$
So $x^2=-1+i\;\;,-1-i$. Means $x$ must be in the form of $a+ib\;,$ where $a,b\neq 0$ and $i=\sqrt{-1}$
So option $(d)$ mut be Correct.
But my question is how can we solve the given question in Analytical method. means not by guessing.
Help me
Thanks
Best Answer
If $p(r) = 0$ it must be that $p(p(x)) = 0$ when $p(x) = r$ or when $p(x)$ equals one of it's roots.
The only (quadratic) polynomial with purely imaginary roots is $p(x) = ax^2 + b$.
If we let the roots $r = \pm iq$ then the solution to $p(x) = ax^2 + b =\pm iq$ cannot be pure real or pure imaginary.
If $x$ were pure real or pure imaginary the left hand side would be real so there would be no way to solve the equation.